Question

A 5.00 g bullet moving with an initial speed of 340 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 860 N/m.

(a) If the block moves 5.50 cm to the right after impact, find the speed at which the bullet emerges from the block
(b) If the block moves 5.50 cm to the right after impact, find the mechanical energy converted into internal energy in the collision

Don't you need the (1/2)kx^2 for the final part of the momentum? I tried just adding it to the final momentum and solving from there but my answer is way off

Answer 1

If the block moves 5.50 cm, the kinetic energy of the block, after the bullet passes through, is
K.E. = (1/2) k X^2 = 430*(0.055)^2
= 1.30 J = (1/2) M V^2

If M and V are the block's mass and velocity, M V = sqrt (2*M*K.E.) = 1.14 kg m/s

That equals the momentum lost by the bullet, m(v - v'). v' is the velcoity after emerging from the block.
1.14 = 0.005*(340 - v')
340 - v' = 228
v' = 112 m/s

For part (b), just do the bookkeeping. You know the initial and final velocities and can compute the kinetic energies.

KE lost = (initial bullet KE) - (final bullet KE) - (final block KE)

Answer 2

i tried those answers and they don't work

Answer 3

i figured it out. thanks for your help!

Answer 4

i do know